Precalculus Examples

$f (x) = x^{2} - 10 x + 5$ ; interval $- 4$ , $4$

Step 1

Write $f (x) = x^{2} - 10 x + 5$ as an equation.

$y = x^{2} - 10 x + 5$

Step 2

Substitute using the average rate of change formula.

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Step 2.1

The average rate of change of a function can be found by calculating the change in $y$ values of the two points divided by the change in $x$ values of the two points.

$\frac{f (4) - f (- 4)}{(4) - (- 4)}$

Step 2.2

Substitute the equation $y = x^{2} - 10 x + 5$ for $f (4)$ and $f (- 4)$ , replacing $x$ in the function with the corresponding $x$ value.

$\frac{({(4)}^{2} - 10 \cdot 4 + 5) - ({(- 4)}^{2} - 10 \cdot - 4 + 5)}{(4) - (- 4)}$

Step 3

Simplify the expression.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $4$ to the power of $2$ .

$\frac{16 - 10 \cdot 4 + 5 - ({(- 4)}^{2} - 10 \cdot - 4 + 5)}{4 - (- 4)}$

Step 3.1.2

Multiply $- 10$ by $4$ .

$\frac{16 - 40 + 5 - ({(- 4)}^{2} - 10 \cdot - 4 + 5)}{4 - (- 4)}$

Step 3.1.3

Simplify each term.

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Step 3.1.3.1

Raise $- 4$ to the power of $2$ .

$\frac{16 - 40 + 5 - (16 - 10 \cdot - 4 + 5)}{4 - (- 4)}$

Step 3.1.3.2

Multiply $- 10$ by $- 4$ .

$\frac{16 - 40 + 5 - (16 + 40 + 5)}{4 - (- 4)}$

Step 3.1.4

Add $16$ and $40$ .

$\frac{16 - 40 + 5 - (56 + 5)}{4 - (- 4)}$

Step 3.1.5

Add $56$ and $5$ .

$\frac{16 - 40 + 5 - 1 \cdot 61}{4 - (- 4)}$

Step 3.1.6

Multiply $- 1$ by $61$ .

$\frac{16 - 40 + 5 - 61}{4 - (- 4)}$

Step 3.1.7

Subtract $40$ from $16$ .

$\frac{- 24 + 5 - 61}{4 - (- 4)}$

Step 3.1.8

Add $- 24$ and $5$ .

$\frac{- 19 - 61}{4 - (- 4)}$

Step 3.1.9

Subtract $61$ from $- 19$ .

$\frac{- 80}{4 - (- 4)}$

Step 3.2

Simplify the denominator.

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Step 3.2.1

Multiply $- 1$ by $- 4$ .

$\frac{- 80}{4 + 4}$

Step 3.2.2

Add $4$ and $4$ .

$\frac{- 80}{8}$

Step 3.3

Divide $- 80$ by $8$ .

$- 10$